Get week number in month from date in PHP?

问题

I have an array of random dates (not coming from MySQL). I need to group them by the week as Week1, Week2, and so on upto Week5.

What I have is this:

$dates = array(\'2015-09-01\',\'2015-09-05\',\'2015-09-06\',\'2015-09-15\',\'2015-09-17\');

What I need is a function to get the week number of the month by providing the date.

I know that I can get the weeknumber by doing date(\'W\',strtotime(\'2015-09-01\')); but this week number is the number between year (1-52) but I need the week number of the month only, e.g. in Sep 2015 there are 5 weeks:

Week1 = 1st to 5th
Week2 = 6th to 12th
Week3 = 13th to 19th
Week4 = 20th to 26th
Week5 = 27th to 30th

I should be able to get the week Week1 by just providing the date e.g.

$weekNumber = getWeekNumber(\'2015-09-01\') //output 1;
$weekNumber = getWeekNumber(\'2015-09-17\') //output 3;

回答1:

I think this relationship should be true and come in handy:

Week of the month = Week of the year - Week of the year of first day of month + 1

Implemented in PHP you get this:

function weekOfMonth($date) {
    //Get the first day of the month.
    $firstOfMonth = strtotime(date("Y-m-01", $date));
    //Apply above formula.
    return intval(date("W", $date)) - intval(date("W", $firstOfMonth)) + 1;
}

To get weeks that starts with sunday, simply replace both date("W", ...) with strftime("%U", ...).

回答2:

You can use the function below, fully commented:

/**
 * Returns the number of week in a month for the specified date.
 *
 * @param string $date
 * @return int
 */
function weekOfMonth($date) {
    // estract date parts
    list($y, $m, $d) = explode('-', date('Y-m-d', strtotime($date)));

    // current week, min 1
    $w = 1;

    // for each day since the start of the month
    for ($i = 1; $i <= $d; ++$i) {
        // if that day was a sunday and is not the first day of month
        if ($i > 1 && date('w', strtotime("$y-$m-$i")) == 0) {
            // increment current week
            ++$w;
        }
    }

    // now return
    return $w;
}

回答3:

The corect way is

function weekOfMonth($date) {
    $firstOfMonth = date("Y-m-01", strtotime($date));
    return intval(date("W", strtotime($date))) - intval(date("W", strtotime($firstOfMonth)));
}

回答4:

I have created this function on my own, which seems to work correctly. In case somebody else have a better way of doing this, please share.. Here is what I have done.

function weekOfMonth($qDate) {
    $dt = strtotime($qDate);
    $day  = date('j',$dt);
    $month = date('m',$dt);
    $year = date('Y',$dt);
    $totalDays = date('t',$dt);
    $weekCnt = 1;
    $retWeek = 0;
    for($i=1;$i<=$totalDays;$i++) {
        $curDay = date("N", mktime(0,0,0,$month,$i,$year));
        if($curDay==7) {
            if($i==$day) {
                $retWeek = $weekCnt+1;
            }
            $weekCnt++;
        } else {
            if($i==$day) {
                $retWeek = $weekCnt;
            }
        }
    }
    return $retWeek;
}

echo weekOfMonth('2015-09-08') // gives me 2;

回答5:

function getWeekOfMonth(DateTime $date) {
    $firstDayOfMonth = new DateTime($date->format('Y-m-1'));

    return ceil(($firstDayOfMonth->format('N') + $date->format('j') - 1) / 7);
}

Goendg solution does not work for 2016-10-31.

回答6:

Given the time_t wday (0=Sunday through 6=Saturday) of the first of the month in firstWday, this returns the (Sunday-based) week number within the month:

weekOfMonth = floor((dayOfMonth + firstWday - 1)/7) + 1

Translated into PHP:

function weekOfMonth($dateString) {
  list($year, $month, $mday) = explode("-", $dateString);
  $firstWday = date("w",strtotime("$year-$month-1"));
  return floor(($mday + $firstWday - 1)/7) + 1;
}

回答7:

function weekOfMonth($strDate) {
  $dateArray = explode("-", $strDate);
  $date = new DateTime();
  $date->setDate($dateArray[0], $dateArray[1], $dateArray[2]);
  return floor((date_format($date, 'j') - 1) / 7) + 1;  
}

weekOfMonth ('2015-09-17') // returns 3

回答8:

You can also use this simple formula for finding week of the month

$currentWeek = ceil((date("d",strtotime($today_date)) - date("w",strtotime($today_date)) - 1) / 7) + 1;

ALGORITHM :

Date = '2018-08-08' => Y-m-d

Find out day of the month eg. 08
Find out Numeric representation of the day of the week minus 1 (number of days in week) eg. (3-1)
Take difference and store in result
Subtract 1 from result
Divide it by 7 to result and ceil the value of result
Add 1 to result eg. ceil(( 08 - 3 ) - 1 ) / 7) + 1 = 2

回答9:

My function. The main idea: we would count amount of weeks passed from the month's first date to current. And the current week number would be the next one. Works on rule: "Week starts from monday" (for sunday-based type we need to transform the increasing algorithm)

function GetWeekNumberOfMonth ($date){
    echo $date -> format('d.m.Y');
    //define current year, month and day in numeric
    $_year = $date -> format('Y');
    $_month = $date -> format('n');
    $_day = $date -> format('j');
    $_week = 0; //count of weeks passed
    for ($i = 1; $i < $_day; $i++){
        echo "\n\n-->";
        $_newDate = mktime(0,0,1, $_month, $i, $_year);
        echo "\n";
        echo date("d.m.Y", $_newDate);
        echo "-->";
        echo date("N", $_newDate);
        //on sunday increasing weeks passed count
        if (date("N", $_newDate) == 7){
            echo "New week";
            $_week += 1;
        }

    }
    return $_week + 1; // as we are counting only passed weeks the current one would be on one higher
}

$date = new DateTime("2019-04-08");
echo "\n\nResult: ". GetWeekNumberOfMonth($date);

回答10:

// self::DAYS_IN_WEEK = 7;
function getWeeksNumberOfMonth(): int
{
    $currentDate            = new \DateTime();
    $dayNumberInMonth       = (int) $currentDate->format('j');
    $dayNumberInWeek        = (int) $currentDate->format('N');
    $dayNumberToLastSunday  = $dayNumberInMonth - $dayNumberInWeek;
    $daysCountInFirstWeek   = $dayNumberToLastSunday % self::DAYS_IN_WEEK;
    $weeksCountToLastSunday = ($dayNumberToLastSunday - $daysCountInFirstWeek) / self::DAYS_IN_WEEK;

    $weeks = [];
    array_push($weeks, $daysCountInFirstWeek);
    for ($i = 0; $i < $weeksCountToLastSunday; $i++) {
        array_push($weeks, self::DAYS_IN_WEEK);
    }
    array_push($weeks, $dayNumberInWeek);

    if (array_sum($weeks) !== $dayNumberInMonth) {
        throw new Exception('Logic is not valid');
    }

    return count($weeks);
}

Short variant:

(int) (new \DateTime())->format('W') - (int) (new \DateTime('first day of this month'))->format('W') + 1;

回答11:

There is a many solutions but here is one my solution that working well in the most cases.

function current_week ($date = NULL) {
    if($date) {
        if(is_numeric($date) && ctype_digit($date) && strtotime(date('Y-m-d H:i:s',$date)) === (int)$date)
            $unix_timestamp = $date;
        else
            $unix_timestamp = strtotime($date);
    } else $unix_timestamp = time();

    return (ceil((date('d', $unix_timestamp) - date('w', $unix_timestamp) - 1) / 7) + 1);
}

It accept unix timestamp, normal date or return current week from the time() if you not pass any value.

Enjoy!

回答12:

//It's easy, no need to use php function
//Let's say your date is 2017-07-02

$Date = explode("-","2017-07-02");
$DateNo = $Date[2];
$WeekNo = $DateNo / 7; // devide it with 7
if(is_float($WeekNo) == true)
{
   $WeekNo = ceil($WeekNo); //So answer will be 1
}  

//If value is not float then ,you got your answer directly

来源：https://stackoverflow.com/questions/32615861/get-week-number-in-month-from-date-in-php

标签

php

arrays

date

week-number